## Accenture Sample Problems On Speed

Below are three problems dealing with speed and time calculations.

Question 1
Two buses leaving from two stations 20 km away from each other travel with constant speed of 45km/hr towards each other.Find the time taken to cross each other if the length of each bus is 100m.
a)10sec b)4sec c)15sec d)20sec
Solution :
The time taken to cross each other = (a + b) / (u + v) sec.
Here, a = b = length of the two buses = 100m = 1/10 km.
and u = v = speed of the two buses = 45km/hr.
Time taken to cross each other = (a + b) / (u + v) sec = (1/10 + 1/10) / (45 + 45) hours
1 / (10 x 90) = 1/900 hours.
since 1 hour = 3600 sec, 1/900 hour = 3600/900 = 4 sec
Hence the answer is 4 sec.

Question 2
A bus starts from A at 7 a.m and reaches the destination B at 7.30 a.m while a cyclist starts from B at 7 a.m and reaches A at 8.30 a.m. At what time the bus and the cyclist will cross each other?
a)7.34 a.m b)7.49 a.m c)7.23 a.m d)8.01 a.m
Solution :
Let the distance between A and B is X km
Time taken by the bus to cover X km = 1/2 hour .(i.e., 7 a.m to 7.30 a.m)
Time taken by the cyclist to cover X km = 3/2 hour (i.e., 7 a.m to 8.30 a.m)
Speed = distance / time
Speed of the bus = X / 1/2 = 2X km/hr
And the speed of the cyclist = X / 3/2 = 2X / 3 km/hr.
Let they meet at point C at Y hours after 7 a.m.
Distance covered by the bus in Y hours = Speed X Time = 2XY km = AC
And the distance covered by the cyclist in Y hours = Speed x Time = 2XY/3 km = BC
Since C is the point where Bus and Cyclist cross each other, AC + BC = AB
In other words, 2XY + (2/3)XY = X
Y(2 + 2/3) = 1
Y = 3/8 hours.
Expressing in minutes, Y = 3/8 x 60 minutes = 22.5 minutes
Y = 23 minutes (approximately)
Therefore they meet at 7 a.m + 23 minutes = 7.23 a.m
Hence the answer is 7.23 a.m.

Question 3
A man rides a bike with a constant speed of 20 km/hr from A at 4 p.m and travels towards a destination B. Another rider with a speed of 25 km/hr starts from B at 5 p.m and travels towards A. If the straight distance between A and B is 110km then at what time they will meet?
a)7 p.m b)8 p.m c)7.30 p.m d)6 p.m
Answer : a)7 p.m
Solution :
Let the two riders meet at X hours after 4 p.m.
Speed of the first rider is 20km/hr and the distance covered by him in X hours = 20X km
Speed of the 2nd rider is 25 Km/hr and he starts at 5 p.m.
Distance covered by him in (X - 1) hours = 25(X - 1) km
Distance between A and B is 110km.
Therefore 20X + 25(X - 1) = 110
45X = 135
X = 135/45 = 3
So they meet at 3 hours after 4 p.m.
Hence the answer is 7 p.m.

## Wipro Sample Algebraic Problems

Below are three model problems dealing with the formulae (x+y)2 = x2 + y2 + 2xy and (x-y)2 = x2 + y2 - 2xy
Question 1
Find X when X - Y = 3 and X2 + Y2 = 89 where X and Y are integers.
a)10 b)-5 c)-10 d)-3
Solution :
We know that (x - y)2 = x2 + y2 - 2xy
Sub. the given values,
32 = 89 - 2XY
9 - 89 = -2XY
80 = 2XY
XY = 40
Since X and Y are integers and XY = 40,the possibilities of X and Y are as follows:
(1,40), (2,20), (4,10), (5,8), (-1,-40), (-2,-20), (-4,-10) and (-5,-8)
X - Y is a positive integer(3), so X will be greater than Y.
By checking the given condition X - Y = 3, we have X = 8, Y = 5 or X = -5, Y = -8.
i.e., X is either 8 or -5.
From the given options we can conclude that X = -5.
Question 2
If the summation and multiplication of two integer is 24 and 143 respectively then the difference of them is:
a)2 b)1 c)12 d)4
Solution :
Let A and B be two integers.
Then A + B = 24 and AB = 143.
We know that (x+y)2 = x2 + y2 + 2xy
Here, 242 = A2 + B2 + 2(143)
576 - 286 = A2 + B2
A2 + B2 = 290.
i.e., Summation of the square of two integers is 290.
i.e., A2 or B2 is < or = 290.
Since A and B are integers, the possibilities are {1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289}.
From the above set of square numbers, the values may be
A2 + B2 = 1 + 289 or 121 + 169 = 290
A2 + B2 = 12 + 172 or 112 + 132 = 290
The condition A + B = 24 and AB = 143, is satisfied by A = 11, B = 13 or A = 13, B = 11
We have to find the difference of A and B.
Hence the difference is 2.
Question 3
If X - Y = 9, X2 + Y2 = 257 and X,Y are integers then what will be the value of X and Y ?
a)21,12 b)15,6 c)none of these d)cannot be determined
Answer : d)cannot be determined
Solution :
Here we use the formula, (x-y)2 = x2 + y2 - 2xy
92 = 257 - 2XY
XY = (257-81)/2 = 176/2 = 88
XY = 88
Since X and Y are integers and XY = 88 then the possibilities of X and Y are (1,88), (2,44), (4,22), (8,11), (-1,-88), (-2,-44), (-4,-22) and (-8,-11).
By observing, none of the above possibility of X and Y satisfies X - Y = 9.
Hence, the values of X and Y cannot be determined by given conditions.

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